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Question-165942




Question Number 165942 by Bagus1003 last updated on 10/Feb/22
Commented by naka3546 last updated on 10/Feb/22
Properties  of  a triangle  a + b > c  a + c > b  b + c > a    1+2 > 3 ?
$$\mathrm{Properties}\:\:\mathrm{of}\:\:\mathrm{a}\:\mathrm{triangle} \\ $$$${a}\:+\:{b}\:>\:{c} \\ $$$${a}\:+\:{c}\:>\:{b} \\ $$$${b}\:+\:{c}\:>\:{a} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{2}\:>\:\mathrm{3}\:? \\ $$
Commented by ajfour last updated on 10/Feb/22
wrong diagram
$${wrong}\:{diagram} \\ $$
Commented by Bagus1003 last updated on 10/Feb/22
thats correct, the heron′s rule is wrong
$${thats}\:{correct},\:{the}\:{heron}'{s}\:{rule}\:{is}\:{wrong} \\ $$
Commented by MJS_new last updated on 10/Feb/22
no. Heron′s Rule is not wrong. but it only  applies to triangles. even in this special case  of a degenerated triangle it′s true: the area  of a triangle is (L/4) which indeed is zero here.
$$\mathrm{no}.\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Rule}\:\mathrm{is}\:\mathrm{not}\:\mathrm{wrong}.\:\mathrm{but}\:\mathrm{it}\:\mathrm{only} \\ $$$$\mathrm{applies}\:\mathrm{to}\:\mathrm{triangles}.\:\mathrm{even}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{case} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{degenerated}\:\mathrm{triangle}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}:\:\mathrm{the}\:\mathrm{area} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{is}\:\frac{{L}}{\mathrm{4}}\:\mathrm{which}\:\mathrm{indeed}\:\mathrm{is}\:\mathrm{zero}\:\mathrm{here}. \\ $$
Commented by Bagus1003 last updated on 11/Feb/22
correct
$${correct} \\ $$
Answered by alephzero last updated on 11/Feb/22
This is right.  Draw one side AB then take a  compass and from each point draw   a circle whose radius will be equal  to your desired side. Then  intersection of these circles will be  point C.
$${This}\:{is}\:{right}. \\ $$$${Draw}\:{one}\:{side}\:{AB}\:{then}\:{take}\:{a} \\ $$$${compass}\:{and}\:{from}\:{each}\:{point}\:{draw}\: \\ $$$${a}\:{circle}\:{whose}\:{radius}\:{will}\:{be}\:{equal} \\ $$$${to}\:{your}\:{desired}\:{side}.\:{Then} \\ $$$${intersection}\:{of}\:{these}\:{circles}\:{will}\:{be} \\ $$$${point}\:{C}. \\ $$
Commented by alephzero last updated on 11/Feb/22
$$ \\ $$$$ \\ $$
Commented by alephzero last updated on 11/Feb/22
$$ \\ $$$$ \\ $$
Commented by alephzero last updated on 11/Feb/22
A_△  = 0
$${A}_{\bigtriangleup} \:=\:\mathrm{0} \\ $$

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